Difference between revisions of "Passing to the quotient (topology)"

Revision as of 12:51, 27 April 2016

Statement

Suppose that [ilmath](X,\mathcal{ J })[/ilmath] is a topological space and [ilmath]\sim[/ilmath] is an equivalence relation, let [ilmath](\frac{X}{\sim},\mathcal{ Q })[/ilmath] be the resulting quotient topology and [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath] the resulting quotient map, then:

• Let [ilmath](Y,\mathcal{ K })[/ilmath] be any topological space and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map that is constant on the fibres of [ilmath]\pi[/ilmath]^{[Note 1]} then:
• there exists a unique continuous map, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] such that [ilmath]f=\overline{f}\circ\pi[/ilmath]

We may then say [ilmath]f[/ilmath] descends to the quotient or passes to the quotient

Note: this is an instance of passing-to-the-quotient for functions

Proof

References

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